Question

If $\omega$ and ${\omega }^2$ are the two imaginary cube roots of unity, then the equation, whose roots are $a{\omega }^{317}$ and $a{\omega }^{382}$, is

• A. $x^2+a^{2}x+a^2=0$
• B. $x^2+a^{2}x+a=0$
• C. $x^2-ax+a^2=0$
• D. $x^2-a^{2}x+a=0$

Answer 1

The correct option is A

$x^2+a^{2}x+a^2=0$

Step 1 : Information from the question

Since, the two imaginary cube roots of unity are $\omega \text{and}{\omega }^2$,

$$\begin{gathered} \therefore {\omega }^2+\omega +1=0 \\ \Rightarrow {\omega }^2+\omega =-1\dots \left(1\right) \end{gathered}$$

Step 2 : Determination of the equation

When two roots of the equation are given, then we can construct the equation by the formula,

$x^2-\left(\text{sum of roots}\right)x+\text{product of roots}=0$

The sum of roots will be $a{\omega }^{317}+a{\omega }^{382}$,

$\begin{array}{rcl}a{\omega }^{317}+a{\omega }^{382}& =& a\left({\omega }^{317}+{\omega }^{382}\right)\\ & =& a\left(\omega +{\omega }^2\right)\\ & =& -a\end{array}$

Now, the product of roots will be,

$\begin{array}{rcl}a{\omega }^{317}·a{\omega }^{382}& =& a^2\left({\omega }^{317}·{\omega }^{382}\right)\\ & =& a^2\end{array}$

Apply the value of the sum of roots and the product of roots in the above equation,

$$\begin{gathered} \therefore x^2-\left(-a\right)x+a^2=0 \\ \Rightarrow x^2+ax+a^2=0 \end{gathered}$$

Hence, the correct option is (A).